1. Introduction
In this paper, we consider the continuous random energy model (CREM) introduced by Bovier and Kurkova in [Reference Bovier and Kurkova3] based on previous work by Derrida and Spohn [Reference Derrida and Spohn5]. The model is defined as follows. Let
$N\in\mathbb{N}$
. Denote by
$\mathbb{T}_N$
the binary tree with depth N. Given
$u\in\mathbb{T}_N$
, we denote by
$|u|$
the depth of u, and we write
$w\leq u$
if w is an ancestor of u. For all
$u,v\in\mathbb{T}_N$
, let
$u\wedge v$
be the most recent common ancestor of u and v. The CREM is a centered Gaussian process
$(X_u)_{u\in\mathbb{T}_N}$
indexed by the binary tree
$\mathbb{T}_N$
of depth N with covariance function
\begin{align*} \mathbb{E}[X_vX_w] = N \cdot A\bigg(\frac{|v\wedge w|}{N}\bigg) \quad \text{for all } v,w \in \mathbb{T}_N.\end{align*}
To study the CREM, one of the key quantities is the partition function defined as
\begin{equation*} Z_{\beta,N} \,:\!=\, \sum_{|u|=N} {\mathrm{e}}^{\beta X_u}.\end{equation*}
In this paper, we study the negative moments of the partition function, which gives us information on the small values of the partition function. This type of study was conducted in the context of other related models such as homogeneous branching random walks, multiplicative cascades, or Gaussian multiplicative chaos, and came with the name of negative moments, left-tail behavior, small deviations of the partition function. We give a survey of these results in Section 1.2.
1.1. Main result
For this paper, we require that the function A satisfies the following assumption.
Assumption 1.1. We suppose that A is a non-decreasing function defined on the interval [0, 1] such that
$A(0)=0$
and
$A(1)=1$
. Let
$\hat{A}$
be the concave hull of A, and we denote by
$\hat{A}'$
the right derivative of
$\hat{A}$
. Throughout this paper, we assume the following regularity conditions.
-
(i) The function A is differentiable in a neighborhood of 0, i.e. there is an
$x_0\in (0,1]$
such that A is differentiable on the interval
$(0,x_0)$
. Furthermore, we assume that A has a finite right derivative at 0. -
(ii) There exists
$\alpha\in (0,1)$
such that the derivative of A is locally Hölder continuous with exponent
$\alpha$
in a neighborhood of 0, i.e. there exists
$x_1\in (0,1]$
such that
\begin{align*}\sup\limits_{x,y\in [0,x_1],\,x\neq y} \frac{|A'(x)-A'(y)|}{{|x-y|}^\alpha} < \infty.\end{align*}
-
(iii) The right derivative of
$\hat{A}$
at
$x=0$
is finite, i.e.
$\hat{A}'(0) < \infty$
.
Remark 1.1. Points (i)–(iii) are technical conditions that are required in the moment estimates perform in Sections 2.2 and 2.3 so that the estimates only depend on
$\hat{A}'(0)$
.
The free energy of the CREM is defined as
$F_\beta \,:\!=\, \lim_{N\rightarrow\infty}\mathbb{E}[\log Z_{\beta,N}]\mathbb{N}$
, and admits an explicit expression: for all
$\beta\geq 0$
,
\begin{equation} F_\beta = \int_0^1 f\big(\beta\sqrt{\hat{A}'(t)}\big) \mathrm{d}{t}, \quad \text{where} \quad f(x) \,:\!=\, \begin{cases} \displaystyle \frac{x^2}{2} + \log 2, & x< \sqrt{2\log 2}, \\[6pt] \sqrt{2\log 2}x, & x\geq \sqrt{2\log 2}. \end{cases}\end{equation}
This formula was proved in [Reference Bovier and Kurkova3, Theorem 3.3], based on a Gaussian comparison lemma [Reference Bovier and Kurkova3, Lemma 3.2] and previous work in [Reference Capocaccia, Cassandro and Picco4]. While [Reference Bovier and Kurkova3] assumed the function A to be piecewise smooth (line 4 after [Reference Bovier and Kurkova3, (1.2)]) to simplify the article, the proof of [Reference Bovier and Kurkova3, Theorem 3.3] does not use this regularity assumption and the proof also holds for the class of functions that are non-decreasing on the interval [0, 1] such that
$A(0)=0$
and
$A(1)=1$
.
The same paper also showed that the maximum of the CREM satisfies
\begin{equation*} \lim_{N\rightarrow\infty}\frac{1}{N}\mathbb{E}\Big[\max_{|u|=N} X_u\Big] = \sqrt{2\log 2}\int_0^1\sqrt{\hat{A}'(t)}\,\mathrm{d}{t}.\end{equation*}
Combining this with (1.1) indicates that there exists
\begin{equation} \beta_\mathrm{c} \,:\!=\, \frac{\sqrt{2\log2}}{\sqrt{\hat{A}'(0)}} \end{equation}
such that the following phase transition occurs. (i) For all
$\beta< \beta_\mathrm{c}$
, the main contribution to the partition function comes from an exponential number of particles. (ii) For all
$\beta>\beta_\mathrm{c}$
, the maximum starts to contribute significantly to the partition function. The quantity
$\beta_\mathrm{c}$
is sometimes referred to as the static critical inverse temperature of the CREM. In the following, we refer to the subcritical regime
$\beta<\beta_\mathrm{c}$
as the high-temperature regime.
Our goal is to study the negative moments of the partition function
$Z_{\beta,N}$
in the high-temperature regime, and we obtain the following result.
Theorem 1.1. Suppose Assumption 1.1 is true. Let
$\beta<\beta_\mathrm{c}$
. For all
$s>0$
, there exist
$N_0 = N_0(A,\beta,s)\in\mathbb{N}$
and a constant
$C=C(A,\beta,s)$
, independent of N, such that, for all
$N\geq N_0$
,
$\mathbb{E}[(Z_{\beta,N})^{-s}] \leq C\mathbb{E}[Z_{\beta,N}]^{-s}$
.
Remark 1.2. For all
$\beta>0$
,
$N\in\mathbb{N}$
, and
$s>0$
, we have the trivial lower bound provided by the Jensen inequality and the convexity of
$x\mapsto x^{-s}$
:
$\mathbb{E}[(Z_{\beta,N})^{-s}] \geq \mathbb{E}[Z_{\beta,N}]^{-s}$
. Combining this with Theorem 1.1, we see that, for all
$\beta<\beta_\mathrm{c}$
in the high-temperature regime and for all
$s>0$
,
$\mathbb{E}[(Z_{\beta,N})^{-s}]$
is comparable with
$\mathbb{E}[Z_{\beta,N}]^{-s}$
.
1.2. Related models
For the (homogeneous) branching random walk, the typical approach is not to study the partition function directly but to consider the additive martingale
$W_{\beta,N}=Z_{\beta,N}/\mathbb{E}[Z_{\beta,N}]$
, which converges to a pointwise limit
$W_{\beta,\infty}$
. A standard method to establish that
$W_{\beta,\infty}$
has negative moments involves the following observation. Suppose that Y
′, Y
′′, W
′, and W
′′ are independent random variables such that Y
′ and Y
′′ share the law of the increment of the branching random walk, and each of W
′ and W
′′ follows the same distribution as
$W_{\beta,\infty}$
. Then, the limit of the additive martingale
$W_{\beta,\infty}$
satisfies the following fixed point equation in distribution:
\begin{equation} W_{\beta,\infty} \stackrel{(\mathrm{d})}{=} Y'W' + Y''W''. \end{equation}
This is a special case of the so-called smoothing transform, which has been studied extensively in the literature. In the context of multiplicative cascades, which is an equivalent model of the branching random walk, [Reference Molchan12] showed that if
$\mathbb{E}[(Y')^{-s}]<\infty$
for some
$s>0$
, then
$\mathbb{E}[(W_{\beta,\infty})^{-s}]<\infty$
. This result was subsequently extended in [Reference Liu10, Reference Liu11]. More recently, [Reference Hu9] studied the small deviation of the maximum of the branching random walk based on the latter result. On the other hand, [Reference Nikula13] provided small deviations for lognormal multiplicative cascades, thereby refining the result in [Reference Molchan12].
In the context of Gaussian multiplicative chaos (GMC), [Reference Garban, Holden, Sepúlveda and Sun8] showed that given a subcritical GMC measure with mild conditions on the base measure, the total mass of this GMC measure has negative moments of all orders. This expanded on the findings of [Reference Duplantier and Sheffield6, Reference Remy14, Reference Robert and Vargas15], with the base measure taken as the Lebesgue measure restricted to some open set, where these works were interested in quantifying the left-tail behavior of the total mass of the GMC near 0. A final remark is that the proof of [Reference Robert and Vargas15, Theorem 3.6], which concerned the negative moment of the total mass of the GMC, was referred to [Reference Barral and Mandelbrot1], which studied a log-Poisson cascade.
1.3. Proof strategy
For the CREM in general,
$W_{\beta,N}$
is not a martingale, so there is no obvious way to show that
$W_{\beta,\infty}$
exists. To prove Theorem 1.1, we adapt the proof of [Reference Benjamini and Schramm2, Lemma A.3], which studied the multiplicative cascade.
While that argument also involved the smoothing transform, it can be adjusted for a general CREM. In particular, we show that for all
$s>0$
, there exist two positive sequences
$\varepsilon_k$
and
$\eta_k$
that both decay double exponentially to 0 as
$k\rightarrow\infty$
such that, for N sufficiently large,
\begin{equation} \mathbb{P}(Z_{\beta,N} \leq \varepsilon_k\mathbb{E}[Z_{\beta,N}]) \leq \eta_k,\end{equation}
and that for all
$k\in[\![ 1, C'\log N]\!]$
, where
$C'>0$
, there exist
$C=C(A,\beta,s)>0$
and
$c=c(A,\beta,s)>0$
such that
\begin{equation} \varepsilon_k^{-s}\eta_k \leq C\exp\big\{{-}c{\mathrm{e}}^{c k}\big\}.\end{equation}
The proof of (1.4) and (1.5) is by establishing an initial left-tail estimate, and then using the branching property to bootstrap the estimate.
2. Initial estimate for the left tail of the partition function
The main goal of this section is to estimate the left tail of
$Z_{\beta,N}$
. The main idea is to adapt (1.3) to the inhomogeneous setting via a recursive argument of
$Z_{\beta,N}$
that we will explain immediately with the following notation.
We denote by
$[\![ m_1,m_2]\!]$
$\{m_1,m_1+1,\ldots,m_2-1,m_2\}$
for two integers
$m-1<m_2$
. Fix
$k\in[\![ 0,N]\!]$
. Define
$(X^{(k)}_u)_{|u|=N-k}$
to be the centered Gaussian process with covariance function
\begin{equation*} \mathbb{E}\big[X^{(k)}_uX^{(k)}_w\big] = N\bigg(A\bigg(\frac{|v\wedge w|+k}{N}\bigg) - A\bigg(\frac{k}{N}\bigg)\bigg).\end{equation*}
We also introduce the corresponding partition function,
$Z^{(k)}_{\beta,N-k} = \sum_{|u|=N-k} {\mathrm{e}}^{\beta X^{(k)}_u}$
. By the covariance function of
$(X^{(k)}_u)_{|u|=N-k}$
and the definition of
$Z^{(k)}_{\beta,N-k}$
, we obtain
\begin{equation} \mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big] = 2^{N-k}\exp\bigg(\frac{\beta^2 N}{2}\bigg(1-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg).\end{equation}
With this notation,
$Z_{\beta,N}$
admits the one-step decomposition
$Z_{\beta,N} = \sum_{|u|=1}{\mathrm{e}}^{\beta X_u}Z^{u}_{\beta,N-1}$
, where
$\big(Z^{u}_{\beta,N}\big)_{|u|}$
are independent and have the same distribution as
$Z^{(1)}_{\beta,N-1}$
. Then, the task of controlling the left tail of
$Z_{\beta,N}$
can be reduced to controlling
$Z^{(1)}_{\beta,N-1}$
and
${\mathrm{e}}^{X_u}$
for
$|u|=1$
. Then, we decompose
$Z^{(1)}_{\beta,N-1}$
in the same manner. We call this procedure the bootstrap argument, which will be made precise in Section 3. The point is that for the bootstrap argument to work, it requires an initial bound, independent of N large and k small, for the left tail of the partition function
$Z^{(k)}_{\beta,N-k}$
in the high-temperature regime. This bound is provided in the following result.
Proposition 2.1. Let
$\beta<\beta_\mathrm{c}$
. Let
$K=\lfloor 2\log_\gamma N \rfloor$
, where
$\gamma\in (11/10,2)$
. Then there exist
$N_0=N_0(A,\beta)\in\mathbb{N}$
and a constant
$\eta_0 = \eta_0(A,\beta) < 1$
such that, for all
$N\geq N_0$
and
$k\in [\![ 0,K]\!]$
,
$\mathbb{P}\big(Z^{(k)}_{\beta,N-k} \leq \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\big) \leq \eta_0$
, where the expectation of
$Z^{(k)}_{\beta,N-k}$
is given in (2.1).
To prove Proposition 2.1, we might want to apply the Paley–Zygmund type argument directly to
$Z^{(k)}_{\beta,N-k}$
. This, however, will not provide the desired result for the whole high-temperature regime due to the correlation of the model. Happily, this can be resolved by introducing a good truncated partition function, which is defined as follows. For all
$|u|=N$
,
$k\in[\![ 0,N]\!]$
,
$a>0$
, and
$b>0$
, define the truncating set
\begin{align*} G_{u,k} \,:\!=\, \big\{\text{for all}\ n\in[\![ 1,N-k]\!] \mid\, & X_w^{(k)} \leq \beta\cdot N\cdot\big(A\big(\tfrac{n+k}{N}\big) - A\big(\tfrac{k}{N}\big)\big) + a\cdot n + b, \\ & \rm{where}\,\textit{w}\, \rm{is \,an\, ancestor\, of \,}\textit{u}\, \rm{with}\, |w|=n \big\}.\end{align*}
Define the truncated partition function
$Z^{(k),\leq}_{\beta,N-k} \,:\!=\, \sum_{|u|=N-k}{\mathrm{e}}^{\beta X^{(k)}_u} \boldsymbol{1}_{G_{u,k}}$
. In Section 2.3, we will prove that the truncated partition has matching squared first moment and second moment. Once we have these two estimates, by passing to a Paley–Zygmund type argument, we can estimate the left tail of the untruncated partition function.
2.1. Many-to-one and many-to-two
To perform the first and second moment estimates, we state and prove the following many-to-one/two lemmas.
Lemma 2.1. (Many-to-one.) Define
$\big(X^{(k)}_n\big)_{n\in [\![ 0,N-k]\!]}$
to be a non-homogeneous random walk such that, for all
$n\in [\![ 0,N-k-1]\!]$
, the increment
$X^{(k)}_{n+1}-X^{(k)}_n$
is distributed as a centered Gaussian random variable with variance
$N(A((n+1)/N)-A(n/N))$
. Let
\begin{equation*} G_{k} = \bigg\{\textit{for all}\ n\in[\![ 1,N-k]\!]\colon X_n^{(k)} \leq \beta\cdot N\cdot\bigg(A\bigg(\frac{n+k}{N}\bigg) - A\bigg(\frac{k}{N}\bigg)\bigg) + a\cdot n + b\bigg\}. \end{equation*}
Then
$\mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big] = 2^{N-k}\mathbb{E}\big[{\mathrm{e}}^{\beta X_{N-k}^{(k)}}\boldsymbol{1}_{G_{k}}\big]$
.
Proof. By the linearity of expectation, we have
$\mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big] = \sum_{|u|=N-k}\mathbb{E}\big[{\mathrm{e}}^{\beta X_{u}^{(k)}}\boldsymbol{1}_{G_{u,k}}\big]$
. On the other hand, for each u with
$|u|=N-k$
, the path
$\big(X_{w}^{(k)}\big)_{w\leq u}$
has the same distribution as
$\big(X^{(k)}_n\big)_{n\in[\![ 0,N-k]\!]}$
, and the proof follows.
Lemma 2.2. (Many-to-two.) Define non-homogeneous random walks
$\big(X^{(k)}_{\ell,n}\big)_{n\in[\![ 0,N-k]\!]}$
and
$\big(\tilde{X}^{(k)}_{\ell,n}\big)_{n\in[\![ 0,N-k]\!]}$
to be such that:
-
• for all
$n\in[\![ 0,N-k-1]\!]$
, the increments
$X^{(k)}_{\ell,n+1}-X^{(k)}_{\ell,n}$
and
$\tilde{X}^{(k)}_{\ell,n+1}-\tilde{X}^{(k)}_{\ell,n}$
are both distributed as a centered Gaussian random variable with variance
$N(A((n+1)/N)-A(n/N))$
; -
•
$X^{(k)}_{\ell,n}=\tilde{X}^{(k)}_{\ell,n}$
for all
$n\in[\![ 0,\ell]\!]$
; and -
•
$\big(X^{(k)}_{\ell,\ell+m}-X^{(k)}_{\ell,\ell}\big)_{m\in[\![ 0,N-k-\ell]\!]}$
and
$\big(\tilde{X}^{(k)}_{\ell,\ell+m}-\tilde{X}^{(k)}_{\ell,\ell}\big)_{m\in[\![ 0,N-k-\ell]\!]}$
are independent.
Let
\begin{align*} G_{\ell,k} & = \bigg\{\textit{for all}\ n\in[\![ 1,N-k]\!] \colon X_{\ell,n}^{(k)} \leq \beta\cdot N\cdot\bigg(A\bigg(\frac{n+k}{N}\bigg) - A\bigg(\frac{k}{N}\bigg)\bigg) + a\cdot n + b \bigg\}, \\ \tilde{G}_{\ell,k} & = \bigg\{\textit{for all}\ n\in[\![ 1,N-k]\!] \colon \tilde{X}_{\ell,n}^{(k)} \leq \beta\cdot N\cdot\bigg(A\bigg(\frac{n+k}{N}\bigg) - A\bigg(\frac{k}{N}\bigg)\bigg) + a\cdot n + b \bigg\}. \end{align*}
Then
\begin{equation*} \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] = 2^{N-k}\mathbb{E}\big[{\mathrm{e}}^{\beta X_{N-k}^{(k)}}\boldsymbol{1}_{G_{k}}\big] + \sum_{\ell=0}^{N-k-1}2^{2(N-k)-\ell-1}\mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}_{\ell,N-k}^{(k)}\right)} \boldsymbol{1}_{G_{\ell,k}\cap \tilde{G}_{\ell,k}}\Big]. \end{equation*}
Proof. By the linearity of expectation, we have the decomposition
\begin{align} \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] & = \sum_{|u|=|w|=N-k}\mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{u}+X_{w}^{(k)}\right)} \boldsymbol{1}_{G_{u}\cap \tilde{G}_{w}}\Big] \nonumber \\ & = \sum_{\ell=0}^{N-k}\,\sum_{\substack{|u|=|w|=N-k,\\|{u\wedge w}|=\ell}} \mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{u}+X_{w}^{(k)}\right)}\boldsymbol{1}_{G_{u}\cap G_{w}}\Big]. \end{align}
For all
$\ell\in[\![ 0,N-k-1]\!]$
, we have
$\big(X^{(k)}_{u_1},X^{(k)}_{u_2}\big)_{u_1\leq u,w\leq u_2} \stackrel{(\mathrm{d})}{=} \big(X^{(k)}_{\ell,n_1},\tilde{X}^{(k)}_{\ell,n_2}\big)_{n_1,n_2\in [\![ 0,N-k]\!]}$
, and for
$\ell= N-k$
,
$\big(X^{(k)}_{u_1}\big)_{u_1\leq w} = \big(X^{(k)}_{u_2}\big)_{u_2\leq w} \stackrel{(\mathrm{d})}{=} \big(X^{(k)}_n\big)_{n\in [\![ 0,N-k]\!]}$
, where
$\big(X^{(k)}_n\big)_{n\in [\![ 0,N-k]\!]}$
was introduced in Lemma 2.1. On the other hand, the number of pairs (u, w) with
$|u|=|w|=N-k$
and
$|{u\wedge w}|=\ell$
is given by
\begin{equation*} \begin{cases} 2^{2(N-k)-\ell-1}, & \ell\in[\![ 0,N-k-1]\!], \\ 2^{N-k}, & \ell=N-k. \end{cases} \end{equation*}
Therefore, (2.2) yields
\begin{equation*} \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] = 2^{N-k}\mathbb{E}\big[{\mathrm{e}}^{\beta X_{N-k}^{(k)}} \boldsymbol{1}_{G_{k}}\big] + \sum_{\ell=0}^{N-k-1}2^{2(N-k)-\ell-1}\mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}_{\ell,N-k}^{(k)}\right)} \boldsymbol{1}_{G_{\ell,k}\cap \tilde{G}_{\ell,k}}\Big], \end{equation*}
and the proof is completed.
2.2. Two useful lemmas for the moment estimates
The following lemma estimates the difference between A(y) and A(z) when y and z are close to 0.
Lemma 2.3. For all
$y,z\in [0,x_1]$
, where
$x_1$
is given in Assumption 1.1(ii), there exists a constant
$C=C(A,x_1)>0$
such that
$|{A\left(y\right) - A\left( z\right)}| \leq \hat{A}'(0) |{y-z}| + Cy^\alpha|{y-z}| + C{|y-z|}^{1+\alpha}$
, where
$\alpha$
appeared in Assumption 1.1(ii).
Proof. Fix
$y,z\in [0,x_1]$
, where
$x_1$
is given in Assumption 1.1(ii). Then, by applying the first-order Taylor expansion of A at y with Lagrange remainder, we have
\begin{equation*} A(z) = A(y) + A'(y)(z-y) + (z-y)(A'(\xi)-A'(y)), \end{equation*}
where
$\xi$
is between z and y. Then, the triangle inequality and the fact that A is non-decreasing yield
\begin{equation} |{A(z)-A(y)}|\leq A'(y)|{z-y}| + |{z-y}||{A'(\xi)-A'(y)}|. \end{equation}
By the local Hölder continuity of A in a neighborhood around 0 provided by Assumption 1.1, there exists
$C=C(A,x_1)>0$
such that
\begin{equation} A'(y)\leq A'(0) + Cy^\alpha. \end{equation}
Similarly, by the local Hölder continuity of A in a neighborhood around 0 provided by Assumption 1.1, we have
\begin{equation} |{A'(\xi)-A'(y)}|\leq C|{z-y}|^\alpha, \end{equation}
where C is the same constant as in (2.4). Combining (2.3)–(2.5), we conclude that
\begin{equation*} |{A(z)-A(y)}|\leq \hat{A}'(0)|{z-y}| + Cy^\alpha|{z-y}| + C|{z-y}|^{1+\alpha}, \end{equation*}
and this completes the proof.
We now state and prove the following exponential tilting formula that we apply later.
Lemma 2.4. (Exponential tilting.) Fix
$m\in\mathbb{N}$
. Let
$(Y_n)_{n\in[\![ 1,m]\!]}$
be a collection of independent centered Gaussian variables with variance
$\sigma_n^2 > 0$
. For all
$n\in[\![ 1,m]\!]$
, define
$S_n = Y_1+\cdots +Y_n$
. Let
$g\colon[\![ 1,m]\!]\rightarrow\mathbb{R}$
be a function, and
$G_n = \{\textit{for all}\ \ell\in[\![ 1,n]\!]\colon S_\ell \leq g(\ell)\}$
,
$n\in[\![ 1,m]\!]$
. Then, for all
$\beta\in\mathbb{R}$
,
$\mathbb{E}\big[{\mathrm{e}}^{\beta S_n}\boldsymbol{1}_{G_n}\big] = \mathbb{E}\big[{\mathrm{e}}^{\beta S_n}\big] \mathbb{P}\big(\textit{for all}\ \ell\in[\![ 1,n]\!]\colon S_\ell \leq g(\ell)-\beta \sum_{r=1}^\ell \sigma_r^2\big)$
.
Proof. Fix
$n\in\mathbb{N}$
. Firstly, from the independence of
$\{Y_i\}_{i=1}^n$
we have
\begin{equation} \mathbb{E}\big[{\mathrm{e}}^{\beta S_n}\big] = \prod_{i=1}^n\mathbb{E}\big[{\mathrm{e}}^{\beta Y_i}\big] = \prod_{i=1}^n\exp\bigg(\frac{\beta^2}{2}\sigma_i^2\bigg) = \exp\bigg(\frac{\beta}{2}\sum_{i=1}^n \sigma_i^2\bigg). \end{equation}
Next, for all
$x_1,\ldots,x_n\in\mathbb{R}$
, we define
\begin{align*} G_n(x_1,\ldots,x_n) & = \{\text{for all}\ \ell\in[\![ 1,n]\!]\colon x_1+\cdots+x_\ell \leq g(\ell)\}, \\ \tilde{G}_n(x_1,\ldots,x_n) & = \big\{\text{for all}\ \ell\in[\![ 1,n]\!]\colon x_1+\cdots+x_\ell \leq g(\ell) - \beta \textstyle\sum_{r=1}^\ell \sigma_r^2\big\}. \end{align*}
Then,
\begin{align*} \mathbb{E}\big[{\mathrm{e}}^{\beta S_n}\boldsymbol{1}_{G_n}\big] & = \frac{1}{\prod_{i=1}^n \sqrt{2\pi \sigma_i^2}\,} \int_{\mathbb{R}^n}{\mathrm{e}}^{\beta\sum_{i=1}^n x_i} \boldsymbol{1}_{G_n(x_1,\ldots,x_n)} \exp\bigg(-\sum_{i=1}^n \frac{x^2_i}{2\sigma_i^2}\bigg) \mathrm{d}{x_1}\cdots\mathrm{d}{x_n} \\ & = \exp\bigg(\frac{\beta^2}{2}\sum_{i=1}^n \sigma_i^2\bigg) \\ & \quad \times \frac{1}{\prod_{i=1}^n\sqrt{2\pi \sigma_i^2}\,} \int_{\mathbb{R}^n}\boldsymbol{1}_{G_n(x_1,\ldots,x_n)} \exp\bigg(-\sum_{i=1}^n \frac{(x_i-\beta\sigma_i^2)^2}{2\sigma_i^2}\bigg) \mathrm{d}{x_1}\cdots\mathrm{d}{x_n} \\ & = \exp\bigg(\frac{\beta^2}{2}\sum_{i=1}^n \sigma_i^2\bigg) \frac{1}{\prod_{i=1}^n \sqrt{2\pi \sigma_i^2}\,} \int_{\mathbb{R}^n} \boldsymbol{1}_{\tilde{G}_n(y_1,\ldots,y_n)} \exp\bigg(\!-\sum_{i=1}^n \frac{y_i^2}{2\sigma_i^2}\bigg) \mathrm{d}{y_1}\cdots\mathrm{d}{y_n} \\ & = \mathbb{E}\big[{\mathrm{e}}^{\beta S_n}\big] \mathbb{P}\Bigg(\text{for all}\ \ell\in[\![ 1,n]\!] \colon S_\ell \leq g(\ell)-\beta \sum_{r=1}^\ell \sigma_r^2\Bigg), \end{align*}
where the second to last line follows from the change of variables
$y_i=x_i - \beta \sigma_i^2$
,
$i=1,\ldots,n$
, and the last line follows from (2.6).
2.3. Moment estimates of the truncated partition function
We now proceed with the first moment estimate.
Lemma 2.5. (First moment estimate.) For all
$a>0$
and
\begin{equation*} b \geq \frac{\hat{A}'(0)}{a}\log\bigg(10\max\bigg\{\frac{{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}}{1-{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}},1\bigg\}\bigg), \end{equation*}
there exists
$N_0=N_0(A,\beta,a,b)\in\mathbb{N}$
such that, for all
$N\geq N_0$
and
$k\in[\![ 0,K]\!]$
where
$K=\lfloor 2\log_\gamma N \rfloor$
with
$\gamma\in(11/10,2)$
,
$\mathbb{E}\big[Z_{\beta,N-k}^{(k),\leq}\big] \geq \frac{7}{10}\mathbb{E}\big[Z_{\beta,N-k}^{(k)}\big]$
.
Proof. By Lemma 2.1, we have
$\mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big] = 2^{N-k}\mathbb{E}\big[e^{\beta X^{(k)}_{N-k}} \boldsymbol{1}_{G_{N-k}}\big]$
. By Lemma 2.4, we have
\begin{multline} \mathbb{E}\big[e^{\beta X^{(k)}_{N-k}} \boldsymbol{1}_{G_{N-k}}\big] \\ = \exp\bigg(\frac{\beta^2 N}{2}\bigg(1-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \cdot \mathbb{P}\big(\text{for all}\ n\in[\![ 1,N-k]\!]\colon X^{(k)}_{N-k}(n)\leq a\cdot n + b \big). \end{multline}
We want to give a lower bound for the probability in (2.7). The union bound and the Chernoff bound yield
\begin{align} & \mathbb{P}\big(\text{there exists}\ n\in[\![ 1,N-k]\!] \colon X^{(k)}_{N-k}(n) > a\cdot n + b\big) \nonumber \\ & \qquad \leq \sum_{n=1}^{N-k}\mathbb{P}\big(X^{(k)}_{N-k}(n)> a\cdot n + b \big) \leq \sum_{n=1}^{N-k}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot (A(({n+k})/{N})-A({k}/{N}))}\bigg). \end{align}
We separate (2.8) into two terms:
\begin{align} (2.8) & = \sum_{n=1}^{K}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \nonumber \\ & \quad + \sum_{n=K+1}^{N-k}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg). \end{align}
We start by bounding the first term of (2.9). Take
$N_1\in\mathbb{N}$
such that
$2K/N\leq x_1$
for all
$N\geq N_1$
. Then, by Lemma 2.3, for all
$n\in [\![ 1,K]\!]$
, we have
\begin{align} N\cdot \bigg(A\bigg(\frac{n+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) & \leq \hat{A}'(0)n + C\frac{n(n+k)^\alpha}{N} + C\frac{n^{1+\alpha}}{N^\alpha} \nonumber \\ & \leq \hat{A}'(0)n + C\frac{K(2K)^\alpha}{N} + C\frac{K^{1+\alpha}}{N^{\alpha}} \end{align}
\begin{align}\quad\quad \quad\quad\, & = \hat{A}'(0) n + h(N) \end{align}
\begin{align}\quad\quad \quad\quad\quad\, & \leq (\hat{A}'(0)+h(N))n, \end{align}
where h(N) is defined as
\begin{equation} h(N)= C(1+2^{\alpha}) \frac{K^{1+\alpha}}{N}. \end{equation}
(2.10) follows by the assumption that
$n\leq K$
and
$k\leq K$
, and (2.12) follows from the fact that
$n\geq 1$
and
$h(N)\geq 0$
.
By (2.12) and the fact that
$(an+b)^2 \geq a^2n^2 + 2abn$
for all
$a,b,n>0$
, the first term of (2.9) is bounded from above by
\begin{align} & \sum_{n=1}^{K}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \nonumber \\ & \qquad \leq \sum_{n=1}^{K}\exp\bigg({-}\frac{a^2 n^2 + 2abn}{2(\hat{A}'(0) + h(N))n}\bigg) \nonumber \\ & \qquad = \exp\bigg(-\frac{ab}{\hat{A}'(0)+h(N)}\bigg)\sum_{n=1}^{K} \exp\bigg(-\frac{a^2n}{2(\hat{A}'(0)+h(N))}\bigg). \end{align}
The summation in (2.14) is bounded from above by
\begin{align*} \sum_{n=1}^{K}\exp\bigg(-\frac{a^2n}{2(\hat{A}'(0)+h(N))}\bigg) & \leq \sum_{n=1}^{\infty}\exp\bigg(-\frac{a^2n}{2(\hat{A}'(0)+h(N))}\bigg) \\ & \leq \frac{\exp\big(\!-{a^2}/({2(\hat{A}'(0)+h(N))})\big)}{1-\exp\big(\!-{a^2}/({2(\hat{A}'(0)+h(N))})\big)}. \end{align*}
This and (2.14) yield
\begin{multline} \sum_{n=1}^{K}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \\ \leq \exp\bigg(-\frac{ab}{\hat{A}'(0)+h(N)}\bigg) \frac{\exp\big(-{a^2}/({2(\hat{A}'(0)+h(N))})\big)}{1-\exp\big(-{a^2}/({2(\hat{A}'(0)+h(N))})\big)}. \end{multline}
Since
$h(N)\rightarrow 0$
as
$N\rightarrow\infty$
, by the choice of parameters a and b we have
\begin{multline*} \limsup_{N\rightarrow\infty}\frac{\exp\big(-{a^2}/({2(\hat{A}'(0)+h(N))})\big)}{1-\exp\big(-{a^2}/({2(\hat{A}'(0)+h(N))})\big)} \exp\bigg(-\frac{ab}{\hat{A}'(0)+h(N)}\bigg) \\ = \frac{{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}}{1-{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}}{\mathrm{e}}^{-ab/\hat{A}'(0)} \leq \frac{1}{10}. \end{multline*}
Combining this with (2.15), there exists
$N_2=N_2(A,\beta)\in\mathbb{N}$
such that, for all
$N\in N_2$
, the first term of (2.9) is bounded from above by
\begin{equation} \sum_{n=1}^{K}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \leq \frac{2}{10}. \end{equation}
It remains to bound the second term of (2.9). By Assumption 1.1, we know that
$A(x)\geq 0$
for all
$x\in [0, 1]$
and
$A(x)\leq \hat{A}(0)x$
for all
$x\in [0, 1]$
. Thus,
\begin{equation} N\cdot\bigg(A\bigg(\frac{n+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq N A\bigg(\frac{n+k}{N}\bigg) \leq \hat{A}'(0)(n+k). \end{equation}
For
$n\geq K+1$
and
$k\leq K$
, we have
$1+{k}/{n} \leq 1+{K}/({K+1}) \leq 2$
. Combining this with (2.17) then yields, for
$n\geq K+1$
and
$k\leq K$
,
\begin{equation*} N\cdot\bigg(A\bigg(\frac{n+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq 2\hat{A}'(0) n. \end{equation*}
Using this and the fact that
$(an+b)^2 \geq a^2 n^2$
for all
$a,b,n>0$
,
\begin{align*} & \sum_{n=K+1}^{N-k}\exp\bigg({-}\frac{(a n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \\ & \qquad\qquad \leq \sum_{n=K+1}^{N-k}\exp\bigg({-}\frac{a^2n^2}{4\hat{A}(0) n}\bigg) \\ & \qquad\qquad = \sum_{n=K+1}^{N-k}\exp\bigg({-}\frac{a^2}{4\hat{A}'(0)}n\bigg) \\ & \qquad\qquad \leq \sum_{n=K+1}^{\infty}\exp\bigg({-}\frac{a^2}{4\hat{A}'(0)}n\bigg) \\ & \qquad\qquad \leq \exp\bigg(-\frac{a^2}{4\hat{A}'(0)}(K+1)\bigg) \frac{1}{1-\exp(\!-{a^2}/({4\hat{A}'(0)}))}. \end{align*}
By our choice of K, the lim sup of this expression equals 0 as
$N\rightarrow\infty$
. In particular, there exists
$N_3=N_3(A,\beta)$
such that, for all
$N\geq N_2$
,
\begin{equation} \sum_{n=K+1}^{N-k}\exp\bigg({-}\frac{(a\cdot n+b)^2}{2N\cdot(A(({n+k})/{N})-A({k}/{N}))}\bigg) \leq \frac{1}{10}. \end{equation}
Let
$N_0\,:\!=\, \max\{N_1,N_2,N_3\}$
. Combining (2.9), (2.16), and (2.18), for all
$N\geq N_0$
and for all
$k\in[\![ 0,K]\!]$
,
\begin{equation} \mathbb{P}\big(\text{there exists}\ n\in[\![ 1,N-k]\!] \colon X^{(k)}_{N-k}(n) > a\cdot n + b\big) \leq \frac{3}{10}. \end{equation}
Combining (2.1), (2.7), (2.7), and (2.19), we conclude that, for all
$N\geq N_0$
and for all
$k\in[\![ 0,K]\!]$
,
\begin{equation*} \mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big] \geq \frac{7}{10}2^{N-k}\exp(\frac{\beta^2 N}{2}\bigg(1-A\bigg(\frac{k}{N}\bigg)\bigg)) = \mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big], \end{equation*}
where the equality follows from (2.1).
It remains to provide a second moment estimate of the truncated partition function.
Lemma 2.6. (Second moment estimate.) Let
$\beta<\beta_\mathrm{c}$
. For all
$a\in\big(0,({\log 2})/{\beta}-\frac{1}{2}\beta\hat{A}'(0)\big)$
and
$b>0$
, there exist
$N_0=N_0(A,\beta)\in\mathbb{N}$
and
$C=C(A,\beta)>0$
such that, for all
$N\geq N_0$
and
$k\in[\![ 0,K ]\!]$
, where
$K=\lfloor 2\log_\gamma N\rfloor$
with
$\gamma\in(11/10,2)$
,
$\mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] \leq C\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]^2$
.
Proof. Fix
$\beta<\beta_\mathrm{c}$
and
$a\in\big(0,({\log 2})/{\beta}-\frac{1}{2}\beta\hat{A}'(0)\big)$
. Define
\begin{equation*} c=c(A,\beta)\,:\!=\, \log 2 - \frac{1}{2}\beta^2\hat{A}'(0)-a >0, \end{equation*}
where the positivity follows from the definition of
$\beta_\mathrm{c}$
in (1.2).
By Lemma 2.2, we have
\begin{align} \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] & = 2^{N-k}\mathbb{E}\big[{\mathrm{e}}^{2\beta X_{N-k}^{(k)}}\boldsymbol{1}_{G_{k}}\big] \nonumber \\ & \quad + \sum_{\ell=0}^{N-k-1}2^{2(N-k)-\ell-1}\mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}_{\ell,N-k}^{(k)}\right)} \boldsymbol{1}_{G_{\ell,k}\cap \tilde{G}_{\ell,k}}\Big]. \end{align}
We now want to estimate the expectations in (2.20). For all
$\ell\in[\![ 1,N]\!]$
and
$n\in[\![ 1, N-\ell]\!]$
, define
\begin{equation*} H_n^{(\ell)} = \bigg\{X_{n}^{(\ell)} \leq \beta N\bigg(A\bigg(\frac{\ell+n}{N}\bigg)-A\bigg(\frac{\ell}{N}\bigg)\bigg) + an + b\bigg\}. \end{equation*}
Then, we see that the expectation in the first term of (2.20) is bounded from above by
\begin{equation*} \mathbb{E}\big[{\mathrm{e}}^{2\beta X_{N-k}^{(k)}}\boldsymbol{1}_{G_{k}}\big] \leq \mathbb{E}\Big[{\mathrm{e}}^{2\beta X^{(k)}_{N-k}}\boldsymbol{1}_{H_{N-k}^{(k)}}\Big]. \end{equation*}
Fix
$\ell\in[\![ 0,N-k-1]\!]$
. We have
\begin{equation} G_{\ell,k}\cap\tilde{G}_{\ell,k} \subseteq \bigg\{X_{\ell,\ell}^{(k)} \leq \beta N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) + a\ell + b\bigg\} = H_{\ell,\ell}^{(k)}. \end{equation}
Then, we have
\begin{align} \mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}_{\ell,N-k}^{(k)}\right)} \boldsymbol{1}_{G_{\ell,k}\cap \tilde{G}_{\ell,k}}\Big] & \leq \mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}_{\ell,N-k}^{(k)}\right)} \boldsymbol{1}_{H_{\ell,\ell}^{(k)}}\Big]\end{align}
\begin{align} & = \mathbb{E}\Big[{\mathrm{e}}^{2\beta X_{\ell,\ell}^{(k)}}\boldsymbol{1}_{H_{\ell,\ell}^{(k)}}\Big] \mathbb{E}\Big[{\mathrm{e}}^{\beta\left(X^{(k)}_{\ell,N-k}+\tilde{X}^{(k)}_{\ell,N-k}-2X^{(k)}_{\ell,\ell}\right)}\Big] \end{align}
\begin{align} & = \mathbb{E}\Big[{\mathrm{e}}^{2\beta X_{\ell}^{(k)}}\boldsymbol{1}_{H_{\ell}^{(k)}}\Big] \mathbb{E}\Big[{\mathrm{e}}^{\beta X^{(\ell+k)}_{N-\ell-k}}\Big]^2 \end{align}
\begin{align} & = \mathbb{E}\Big[{\mathrm{e}}^{2\beta X_{\ell}^{(k)}}\boldsymbol{1}_{H_{\ell}^{(k)}}\Big] \exp\bigg(\beta^2 N\bigg(1-A\bigg(\frac{\ell+k}{N}\bigg)\bigg)\bigg), \end{align}
where (2.22) follows from (2.21), and both (2.23) and (2.24) from Lemma 2.2.
For
$\ell\in[\![ 0,N-k]\!]$
, Lemma 2.4 yields
\begin{align} \mathbb{E}\Big[{\mathrm{e}}^{2\beta X_{\ell}^{(k)}} \boldsymbol{1}_{H_{\ell}^{(k)}}\Big] & = \exp\bigg(2\beta^2 N \bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \nonumber \\ & \quad \times \mathbb{P}\bigg(X_{\ell}^{(k)} \leq -\beta N\bigg(A\bigg(\frac{k+\ell}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) + a \ell + b\bigg). \end{align}
We distinguish the following two cases.
Case 1:
$-\beta N(A(({\ell+k})/{N})-A({k}/{N})) + a \ell + b\geq 0$
. In this case, by bounding the probability in (2.26) by 1, we obtain
\begin{align*} (2.26) & \leq \exp\bigg(2\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \\ & \leq \exp\bigg(\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg)\exp\!(\beta(a\ell+b)). \end{align*}
Case 2:
$-\beta N(A(({\ell+k})/{N})-A({k}/{N})) + a \ell + b<0$
. In this case, the Chernoff bound yields
\begin{align} (2.27) & \leq \exp\bigg(2\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \nonumber \\ & \quad \times \exp\bigg(-\frac{(\beta N(A(({\ell+k})/{N})-A({k}/{N})) - (a \ell+b))^2} {2N(A(({\ell+k})/{N})-A({k}/{N}))}\bigg). \end{align}
By the fact that
\begin{multline*} \bigg(\beta N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) - (a \ell+b)\bigg)^2 \\ \geq \beta^2N^2\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)^2 - 2\beta N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)(a\ell+b), \end{multline*}
(2.27) yields the bound
\begin{equation*} (2.26) \leq \exp\bigg(\frac{3}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta (a \ell + b)). \end{equation*}
From the case study above, we conclude that
\begin{equation} \mathbb{E}\Big[{\mathrm{e}}^{2\beta X_{\ell}^{(k)}} \boldsymbol{1}_{H_{\ell}^{(k)}}\Big] \leq \exp\bigg(\frac{3}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a\ell + b)). \end{equation}
Applying the bounds in (2.25) and (2.28) to (2.20), we obtain
\begin{align} & \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] \nonumber \\ & \leq 2^{N-k}\exp\bigg(\frac{3}{2}\beta^2N\bigg(A\bigg(\frac{1}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a(N-k) + b)) \nonumber \\ & \quad + \sum_{\ell=0}^{N-k-1}2^{2(N-k)-\ell-1}\exp\bigg(\frac{3}{2}\beta^2N \bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg)\exp\!(\beta(a\ell + b)) \nonumber \\ & \quad \times \exp\bigg(\beta^2N\bigg(1-A\bigg(\frac{\ell+k}{N}\bigg)\bigg)\bigg) \nonumber \\ & \leq 2^{2(N-k)}\exp\bigg(\beta^2N\bigg(1-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \nonumber \\ & \quad \times \sum_{\ell=0}^{N-k}2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a\ell + b)) \end{align}
\begin{align} = \mathbb{E}\big[Z_{\beta,N-k}^{(k)}\big]^2\sum_{\ell=0}^{N-k} 2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta (a\ell + b)),\end{align}
where (2.29) follows from (2.1) and the fact that
$2^{-\ell-1}\leq 2^{-\ell}$
for any integer
$\ell$
. Now, we decompose the summation in (2.30) into
\begin{align} & \sum_{\ell=0}^{N-k}2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a\ell + b)) \nonumber \\ & = \sum_{\ell=0}^{K_1}2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a\ell + b)) \nonumber \\ & \quad + \sum_{\ell=K_1+1}^{N-k}2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta(a\ell + b)), \end{align}
where
$K_1=K^2$
. We start by bounding the first term in (2.31). Take
$N_1$
such that
$(K_1+K)/N\leq x_1$
for all
$N\geq N_1$
. Following the same derivation as for (2.11), for all
$N\geq N_1$
,
$k\in[\![ 1,K]\!]$
, and
$\ell\in[\![ 0,K_1]\!]$
, we have
\begin{equation*} N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq \hat{A}'(0)\ell + h(N), \end{equation*}
where h(N) is defined in (2.13). Applying this then yields
\begin{align} \text{first term in (2.31)} & \leq \exp\!(h(N)+\beta b)\sum_{\ell=0}^{K_1}\exp\bigg(\!-\bigg(\log 2-\frac{\beta^2}{2}\hat{A}(0)-\beta a\bigg)\ell\bigg) \nonumber \\ & = \exp\!(h(N)+\beta b)\sum_{\ell=0}^{K_1} \exp\!(\!-c\ell) \nonumber \\ & \leq \exp\!(h(N)+\beta b)\sum_{\ell=0}^{\infty} \exp\!(\!-c\ell) = \exp\!(h(N)+\beta b) \frac{1}{1-{\mathrm{e}}^{-c}}. \end{align}
It remains to bound the second term in (2.31). As in (2.17), Assumption 1.1 implies that
$A(x)\geq 0$
for all
$x\in [0, 1]$
, and
$A(x)\leq \hat{A}(0)x$
for all
$x\in [0, 1]$
, so
\begin{equation*} N\cdot\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq \hat{A}'(0)(\ell+k) \leq \hat{A}'(0)\ell + \hat{A}'(0)K, \end{equation*}
where the second inequality is by the assumption that
$k\leq K$
. Thus,
\begin{align} \text{second term in (2.31)} & \leq \exp(\hat{A}'(0)K+\beta b) \sum_{\ell=K_1+1}^{N-k}\exp\bigg(\!-\bigg(\log 2 - \frac{\beta^2}{2}\hat{A}'(0)-\beta a\bigg)\ell\bigg) \nonumber \\ & = \exp(\hat{A}'(0)K+\beta b)\sum_{\ell=K_1+1}^{N-k}\exp\!(\!-c\ell) \nonumber \\ & \leq \exp(\hat{A}'(0)K+\beta b)\sum_{\ell=K_1+1}^{\infty}\exp\!(\!-c\ell) \nonumber \\ & = \exp(\hat{A}'(0)K+\beta b)\frac{{\mathrm{e}}^{-c(K_1+1)}}{1-{\mathrm{e}}^{-c}} = \exp\!(\!-L(N)+\beta b)\frac{{\mathrm{e}}^{-c}}{1-{\mathrm{e}}^{-c}}, \end{align}
where
$L(N) = c K_1 - \hat{A}'(0)K = c K^2 - \hat{A}'(0)K$
. Combining (2.31), (2.32), and (2.33), we derive that
\begin{multline} \sum_{\ell=0}^{N-k} 2^{-\ell} \exp\bigg(\frac{1}{2}\beta^2N\bigg(A\bigg(\frac{\ell+k}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \exp\!(\beta (a \ell + b)) \\ \leq \exp\!(h(N)+\beta b)\frac{1}{1-{\mathrm{e}}^{-c}} + \exp\!(\!-L(N)+\beta b)\frac{{\mathrm{e}}^{-c}}{1-{\mathrm{e}}^{-c}}. \end{multline}
Note that
$h(N)\rightarrow 0$
and
$L(N)\rightarrow\infty$
as
$N\rightarrow\infty$
, so there exists
$N_2$
such that
$h(N)\leq 1$
and
$L(N)\geq 1$
for all
$N\geq N_2$
. Take
$N_0=\max\{N_1,N_2\}$
. For all
$N\geq N_0$
, we conclude from (2.30) and (2.34) that
$\mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big] \leq C\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]^2$
, where
$C = C(A,\beta) = \exp(1+\beta b)/({1-{\mathrm{e}}^{-c}}) + \exp\!(\!-1+\beta b){{\mathrm{e}}^{-c}}/({1-{\mathrm{e}}^{-c}})$
.
2.4. Proof of Proposition 2.1
We are now ready to prove Proposition 2.1.
Proof of Proposition
2.1. Fix
$\beta<\beta_\mathrm{c}$
. By Lemmas 2.5 and 2.6, if we choose
\begin{equation*} a = \frac{1}{2\beta}\bigg(\log 2-\frac{1}{2}\beta^2\hat{A}'(0)\bigg), \qquad b = \frac{\hat{A}'(0)}{a}\log\bigg(10\max\bigg\{\frac{{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}}{1-{\mathrm{e}}^{-a^2/(2\hat{A}'(0))}},1\bigg\}\bigg), \end{equation*}
then there exist
$N_1=N_1(A,\beta)\in\mathbb{N}$
,
$N_2=N_2(A,\beta)\in\mathbb{N}$
, and
$C=C(A,\beta)>0$
such that, for all
$N\geq N_1$
and
$k\in[\![ 0,K]\!]$
,
\begin{equation} \mathbb{E}\big[Z_{\beta,N-k}^{(k),\leq}\big] \geq \frac{7}{10}\mathbb{E}\big[Z_{\beta,N-k}^{(k)}\big], \end{equation}
and for all
$N\geq N_2$
and
$k\in[\![ 0,K]\!]$
,
\begin{equation} \mathbb{E}\big[\big(Z_{\beta,N-k}^{(k),\leq}\big)^2\big] \leq C \mathbb{E}\big[Z_{\beta,N-k}^{(k)}\big]^2. \end{equation}
Now, since
$Z^{(k),\leq}_{\beta,N-k} \leq Z^{(k)}_{\beta,N-k}$
, we have
\begin{equation} \mathbb{P}\bigg(Z^{(k)}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg) \geq \mathbb{P}\bigg(Z^{(k),\leq}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg). \end{equation}
On the other hand, by the Cauchy–Schwarz inequality, we have
\begin{align} & \mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big] \nonumber \\ & \qquad = \mathbb{E}\bigg[Z^{(k),\leq}_{\beta,N-k} \boldsymbol{1}\bigg\{Z^{(k),\leq}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg\}\bigg] + \underbrace{\mathbb{E}\bigg[Z^{(k),\leq}_{\beta,N-k}\boldsymbol{1}\bigg\{Z^{(k),\leq}_{\beta,N-k} \leq \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg\}\bigg]}_{\leq\frac{1}{2}\mathbb{E}\left[Z^{(k)}_{\beta,N-k}\right]} \nonumber \\ & \qquad \leq \mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big]^{1/2} \mathbb{P}\bigg(Z^{(k),\leq}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg)^{1/2} + \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]. \end{align}
Let
$N_0 = N_0(A,\beta) \,:\!=\, \max\{N_1,N_2\}$
. Combining (2.35), (2.36), and (2.38), we derive that, for all
$N\geq N_0$
and
$k\in[\![ 0,K]\!]$
,
\begin{equation} \mathbb{P}\bigg(Z^{(k),\leq}_{\beta,N-k}\geq\frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg)^{1/2} \geq \frac{\mathbb{E}\big[Z^{(k),\leq}_{\beta,N-k}\big]-\frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]} {\mathbb{E}\big[\big(Z^{(k),\leq}_{\beta,N-k}\big)^2\big]^{1/2}} \geq \frac{2}{10\sqrt{C}\,}. \end{equation}
Combining (2.37) and (2.39), we conclude that, for all
$N\geq N_0$
and
$k\in[\![ 0,K]\!]$
,
\begin{align*} \mathbb{P}\bigg(Z^{(k)}_{\beta,N-k} \leq \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg) & = 1 - \mathbb{P}\bigg(Z^{(k)}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg) \\ & \leq 1 - \mathbb{P}\bigg(Z^{(k),\leq}_{\beta,N-k} > \frac{1}{2}\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\bigg) \leq 1- \frac{4}{100C}. \end{align*}
By taking
$\eta_0 = \eta_0(A,\beta) = 1-{4}/{100 C}$
, the proof is completed.
3. Left tail estimates via a bootstrap argument
The main goal of this section is to provide a finer estimate of the left tail of
$Z_{\beta,N}$
than the one provided by Proposition 2.1. Namely, we want to construct two sequences that satisfy (1.5), and this is manifested in the following proposition.
Proposition 3.1. Let
$K=\lfloor 2 \log_\gamma N \rfloor$
with
$\gamma\in (11/10,2)$
. Then, there exist
$N_0=N_0(A,\beta)\in\mathbb{N}$
and two sequences
$(\eta_k)_{k=0}^K$
and
$(\varepsilon_k)_{k=0}^K$
such that
-
(i) for all
$N\geq N_0$
and
$k\in [\![ 0,K ]\!]$
,
$\mathbb{P}(Z_{\beta,N} \leq \varepsilon_k \mathbb{E}[Z_{\beta,N}]) \leq \eta_k$
; -
(ii) for all
$s>0$
, there exists a constant
$C=C(A,\beta,s)>0$
, independent of K, such that
$\sum_{r=0}^{K-1} (\varepsilon_{r+1})^{-s} \eta_r \leq C$
.
The proof of Proposition 3.1 requires the following two lemmas. For all
$k\in[\![ 0,N-1]\!]$
, define
$M^{(k)}_{\beta,1} = {{\mathrm{e}}^{\beta X^{(k)}_{1}}}/{\mathbb{E}\big[Z^{(k)}_{\beta,1}\big]}$
, where
$X^{(k)}_1$
is defined in Lemma 2.1. The first lemma states a bootstrap inequality.
Lemma 3.1. For all
$c>0$
,
$\delta>0$
,
$N\in\mathbb{N}$
, and
$k\in [\![ 0,N-1]\!]$
,
\begin{equation*} \mathbb{P}\big(Z^{(k)}_{\beta,N-k} \leq c\delta\cdot\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\big) \leq \big[\mathbb{P}\big(Z^{(k+1)}_{\beta,N-k-1} \leq c\cdot\mathbb{E}\big[Z^{(k+1)}_{\beta,N-k-1}\big]\big) + \mathbb{P}\big(M^{(k)}_{\beta,1} \leq \delta\big) \big]^2. \end{equation*}
Proof. Fix
$N\in\mathbb{N}$
. For every
$c>0$
,
$\delta>0$
, and
$k\in [\![ 0,N-1]\!]$
, we have
\begin{align} \mathbb{P}\big(Z^{(k)}_{\beta,N-k} \leq c\delta\cdot\mathbb{E}\big[Z^{(k)}_{\beta,N-k}\big]\big) & = \mathbb{P}\Bigg(\sum_{|u|=1}M_{\beta,u}^{(k)}\cdot Z^{(k),u}_{\beta,N-k-1} \leq c\delta\cdot\mathbb{E}\big[Z^{(k+1)}_{\beta,N-k-1}\big]\Bigg) \nonumber \\ & \leq \mathbb{P}\big(\text{for all}\ |u|=1\colon M_{\beta,u}^{(k)}\cdot Z^{(k),u}_{\beta,N-k-1} \leq c\delta\cdot\mathbb{E}\big[Z^{(k+1)}_{\beta,N-k-1}\big]\big) \nonumber \\ & = \mathbb{P}\big(M_{\beta,1}^{(k)}\cdot Z^{(k+1)}_{\beta,N-k-1} \leq c\delta\cdot\mathbb{E}\big[Z^{(k+1)}_{\beta,N-k-1}\big]\big)^2 \nonumber\\ & \leq \big[\mathbb{P}\big(Z^{(k+1)}_{\beta,N-k-1} \leq c\cdot\mathbb{E}\big[Z^{(k+1)}_{\beta,N-k-1}\big]\big) + \mathbb{P}\big(M_{\beta,1}^{(k)} \leq \delta\big) \big]^2, \end{align}
where (3.1) follows from independence.
The next lemma provides a uniform estimate for the left tail of
$M_{\beta,1}^{(k)}$
when k is small compared to N.
Lemma 3.2. Let
$N\in\mathbb{N}$
and
$K=\lfloor 2 \log_\gamma N \rfloor$
with
$\gamma\in(11/10,2)$
. For all
$\beta>0$
,
$k\in[\![ 0, K]\!]$
, and
$s>0$
, there exist
$N_0=N(A,\beta,s)\in\mathbb{N}$
and a constant
$C=C(A,\beta,s)\geq 1$
, independent of N, such that, for all
$N\geq N_0$
,
$\mathbb{E}\big[(M^{(k)}_{\beta,1})^{-s}\big] \leq C$
. In particular, by Markov’s inequality this implies that, for all
$\delta>0$
and all
$N\geq N_0$
,
$\mathbb{P}\big(M^{(k)}_{\beta,1} \leq \delta\big) \leq C \cdot \delta^s$
.
Proof. Let
$k\in[\![ 0,K]\!]$
. By Lemma 2.3, there exist
$N_1=N_1(A,\beta)\in\mathbb{N}$
and
$C_1>0$
such that, for all
$N\geq N_1$
,
\begin{equation*} N\bigg(A\bigg(\frac{k+1}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq \hat{A}'(0) + C_1\frac{(k+1)^\alpha}{N^\alpha} + C_1\frac{1}{N^\alpha} \leq \hat{A}'(0) + C_1 \frac{(K+1)^\alpha+1}{N^\alpha}. \end{equation*}
Since
$\limsup_{N\rightarrow\infty}C_1({(K+1)^\alpha+1})/{N^\alpha} = 0$
, in particular, there exists
$N_2=N_2(A,\beta)\in\mathbb{N}$
such that, for all
$N\geq N_2$
,
\begin{equation*} N\bigg(A\bigg(\frac{k+1}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg) \leq \hat{A}'(0) + 1. \end{equation*}
Let
$N_0\,:\!=\, \max\{N_1,N_2\}$
and
\begin{equation*} C = C(A,\beta,s) \,:\!=\, \exp\bigg(\frac{\beta^2 s^2}{2} (\hat{A}'(0)+1)\bigg) \cdot 2^s \exp\bigg(\frac{\beta^2 s}{2} (\hat{A}'(0)+1)\bigg). \end{equation*}
Note that this constant is greater than or equal to 1 since
${\beta^2 s^2}(\hat{A}'(0)+1)/{2}$
, s, and
${\beta^2 s}(\hat{A}'(0)+1)/{2}$
are non-negative. For all
$N\geq N_0$
, we conclude that
\begin{align*} & \mathbb{E}\big[(M^{(k)}_{\beta,1})^{-s}\big] \\ & \quad = \mathbb{E}\big[{\mathrm{e}}^{-\beta s X^{(k)}_{1}}\big] \cdot \mathbb{E}\big[Z^{(k)}_{\beta,1}\big]^s \\ & \quad = \exp\bigg(\frac{\beta^2s^2}{2}N\bigg(A\bigg(\frac{k+1}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \cdot 2^s\exp\bigg(\frac{\beta^2s}{2}N\bigg(A\bigg(\frac{k+1}{N}\bigg)-A\bigg(\frac{k}{N}\bigg)\bigg)\bigg) \\ & \quad \leq \exp\bigg(\frac{\beta^2 s^2}{2}(\hat{A}'(0)+1)\bigg) \cdot 2^s \exp\bigg(\frac{\beta^2 s}{2}(\hat{A}'(0)+1)\bigg) = C, \end{align*}
and this completes the proof.
We now turn to the proof of Proposition 3.1.
Proof of Proposition
3.1. Fix
$\gamma \in (11/10,2)$
and
$s>0$
. Let
$\eta_0=\eta(A,\beta)<1$
be the same as in Proposition 2.1, and for all
$k\in[\![ 1,K]\!]$
define
$\eta_k = (\eta_0)^{\gamma^k}$
. Let
$\varepsilon_0=\frac{1}{2}$
. With
$C=C(A,\beta,10s)\geq 1$
being the same constant that appeared in Lemma 3.2, for all
$k\in[\![ 1,K]\!]$
define
$\varepsilon_k$
such that
\begin{equation*} (\varepsilon_k)^s = \frac{1}{2^s}\frac{1}{C^k}\prod_{n=0}^{k-1}\big((\eta_n)^{\gamma/2}-\eta_n\big)^{1/10}. \end{equation*}
We first claim that, for all
$k\in[\![ 0,K]\!]$
,
$\ell\in[\![ 0,K-k]\!]$
, and
$N\geq N_0$
, where
$N_0=N_0(A,\beta)$
appeared in the statement of Proposition 2.1,
\begin{equation} \mathbb{P}\big(Z_{\beta,N-\ell}^{(\ell)} \leq \varepsilon_k\mathbb{E}\big[Z_{\beta,N-\ell}^{(\ell)}\big]\big) \leq \eta_k. \end{equation}
Note that (3.2) implies Proposition 3.1(i) by taking
$\ell=0$
.
We prove (3.2) by induction on k. The case where
$k=0$
follows from Proposition 2.1. Suppose that (3.2) is true for
$k\in[\![ 0,K-1]\!]$
. As
$\varepsilon_{k+1} = \varepsilon_{k}\big(\big(\eta_{k}^{\gamma/2}-\eta_{k}\big)/C\big)^{{1}/{10s}}$
, applying Lemma 3.1 with
$c=\varepsilon_{k}$
and
$\delta=\big(\big(\eta_{k}^{\gamma/2}-\eta_{k}\big)/C\big)^{{1}/{10s}}$
yields that, for all
$\ell\in[\![ 0,K-k-1]\!]$
and
$N\geq N_0$
,
\begin{align*} & \mathbb{P}\big(Z_{\beta,N-\ell}^{(\ell)}\leq\varepsilon_{k+1}\mathbb{E}\big[Z_{\beta,N-\ell}^{(\ell)}\big]\big) \\ & \qquad = \mathbb{P}\bigg(Z_{\beta,N-\ell}^{(\ell)} \leq \varepsilon_{k}\cdot\bigg(\frac{1}{C}\big(\eta_{k}^{\gamma/2}-\eta_{k}\big)\bigg)^{1/s} \mathbb{E}\big[Z_{\beta,N-\ell}^{(\ell)}\big]\bigg) \\ & \qquad \leq \bigg(\mathbb{P}\big(Z_{\beta,N-\ell-1}^{(\ell+1)} \leq \varepsilon_{k}\mathbb{E}\big[Z_{\beta,N-\ell-1}^{(\ell+1)}\big]\big) + \mathbb{P}\bigg(M_{\beta,1}^{(\ell)} \leq \bigg(\frac{1}{C}\big(\eta_{k}^{\gamma/2}-\eta_{k}\big)\bigg)^{{1}/{10s}}\bigg)\bigg)^2 \\ & \qquad \leq \big(\eta_{k}+\eta_{k}^{\gamma/2}-\eta_{k}\big)^2 = (\eta_{k})^{\gamma} = \eta_{k+1}, \end{align*}
where the last inequality follows from the induction hypothesis and Lemma 3.2. Thus, by induction, (3.2) holds for all
$k\in[\![ 0,K]\!]$
.
It remains to show that, for all
$s>0$
,
$\sum_{k=0}^K(\varepsilon_{k+1})^{-s}\eta_{k}$
is uniformly bounded in K. For all
$k\in[\![ 0,K]\!]$
,
\begin{align} (\varepsilon_{k+1})^{-s} & = 2^{s}C^{k+1}\prod_{n=0}^k\big(\eta_n^{\gamma/2}-\eta_n\big)^{-1/10} \nonumber \\ & = 2^{s}C^{k+1}\prod_{n=0}^k\eta_n^{-\gamma/20}\big(1-\eta_n^{1-\gamma/2}\big)^{-(k+1)/10} \nonumber \\ & \leq 2^{s}C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-1/10}\prod_{n=0}^k \eta_n^{-\gamma/20} \end{align}
\begin{align} & = 2^{s}C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-(k+1)/10}\eta_0^{-({\gamma}/{20})(({\gamma^{k+1}-1})/({\gamma-1}))} \nonumber \\ & = C_1 C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-(k+1)/10}\eta_0^{-{\gamma^2\gamma^k}/{20(\gamma-1)}}, \end{align}
where
$C_1=C_1(s,\eta_0,\gamma)=2^{s}\eta_0^{{\gamma}/{20(\gamma-1)}}$
, and (3.3) follows from the definition of
$\eta_n$
and the fact that
$(\eta_k)$
is a decreasing sequence, which implies in particular that
$\eta_k\leq \eta_0$
for all
$k\in[\![ 0,K]\!]$
. Applying (3.4),
\begin{align} \sum_{k=0}^{K-1} (\varepsilon_{k+1})^{-s}\eta_{k} & \leq C_1\sum_{k=0}^{K-1}C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-(k+1)/10} \eta_0^{-{\gamma^2\gamma^k}/{20(\gamma-1)}} \eta_k \nonumber \\ & = C_1\sum_{k=0}^{K-1}C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-(k+1)/10} \eta_0^{(1-{\gamma^2}/{20(\gamma-1)})\gamma^k} \end{align}
\begin{align} & \leq C_1\sum_{k=0}^{\infty}C^{k+1}\big(1-\eta_0^{1-\gamma/2}\big)^{-(k+1)/10} \eta_0^{(1-{\gamma^2}/{20(\gamma-1)})\gamma^k} \nonumber \\ & \leq C_1\sum_{k=0}^\infty\exp\bigg(\!\!\log\bigg(\!\frac{C}{\big(1-\eta_0^{1-\gamma/2}\big)^{1/10}}\bigg)(k+1) + \bigg(\!1-\frac{\gamma^2}{20(\gamma-1)}\bigg)(\!\log\eta_0)\gamma^k\!\bigg),\end{align}
where (3.5) follows from the definition of
$\eta_k$
. Since
$\gamma\in(11/10,2)$
and
$\eta_0 < 1$
,
$1-{\gamma^2}/{20(\gamma-1)}>0$
and
$\log \eta_0 < 1$
. Therefore, (3.6) depends only on s,
$\gamma\in(11/10,2)$
, and
$\eta_0=\eta_0(A,\beta)$
, and we conclude that
$\sum_{k=0}^K (\varepsilon_{k+1})^{-s}\eta_{k}$
is bounded from above by a constant
$C_2=C_2(A,\beta,s)>0$
, as desired.
4. Proof of Theorem 1.1
Proof. Suppose that
$\beta<\beta_\mathrm{c}$
. Let
$N\in\mathbb{N}$
. Fix
$K=\lfloor 2 \log_\gamma N \rfloor$
and
$s>0$
. Also, we define
$W_{\beta,N}\,:\!=\, Z_{\beta,N}/\mathbb{E}[Z_{\beta,N}]$
. It suffices to prove that there exists a constant
$C=C(A,\beta,s)>0$
such that
$\mathbb{E}[(W_{\beta,N})^{-s}] \leq C$
. We have
\begin{align} & \mathbb{E}[(W_{\beta,N})^{-s}] \nonumber \\ & \quad \leq \varepsilon_0^{-s}\mathbb{P}(W_{\beta,N}>\varepsilon_0) + \sum_{k=0}^{K-1}(\varepsilon_{k+1})^{-s}\cdot\mathbb{P}(\varepsilon_{k+1} \leq W_{\beta,N}\leq \varepsilon_k) + \mathbb{E}[(W_{\beta,N})^{-s}\boldsymbol{1}_{W_{\beta,N}\leq \varepsilon_K}] \nonumber \\ & \quad \leq 2^s + \sum_{n=0}^{K-1}(\varepsilon_{k+1})^{-s}\cdot\mathbb{P}(W_{\beta,N}\leq \varepsilon_k) + \mathbb{E}[(W_{\beta,N})^{-s}\boldsymbol{1}_{W_{\beta,N}\leq \varepsilon_K}]. \end{align}
By Proposition 3.1, there exist
$N_1=N(A,\beta,s)\in\mathbb{N}$
and
$C_1=C_1(A,\beta,s)>0$
such that, for all
$N\geq N_1$
, the second term of (4.1) is bounded by
$C_1$
. Thus, it remains to bound the third term of (4.1).
For all
$|u|=N$
,
$Z_{\beta,N} \geq {\mathrm{e}}^{\beta X_u}$
. Fix a vertex u with
$|u|=N$
. By the previous inequality and Lemma 2.4, the second term of (4.1) is bounded from above by
\begin{align} \mathbb{E}[(W_{\beta,N})^{-s}\boldsymbol{1}_{W_{\beta,N}\leq \varepsilon_K}] & \leq \mathbb{E}[Z_{\beta,N}]^{s} \cdot \mathbb{E}\big[({\mathrm{e}}^{\beta X_u})^{-s}\boldsymbol{1}_{{\mathrm{e}}^{\beta X_u} \leq \varepsilon_K\mathbb{E}[Z_{\beta,N}]}\big] \nonumber \\ & = \mathbb{E}[Z_{\beta,N}]^{s} \cdot \mathbb{E}\big[{\mathrm{e}}^{-\beta s X_u}\boldsymbol{1}_{X_u\leq(\log\varepsilon_K+\log\mathbb{E}[Z_{\beta,N}])/{\beta}}\big] \nonumber \\ & = \mathbb{E}[Z_{\beta,N}]^{s} \mathbb{E}\big[e^{-\beta sX_u}\boldsymbol{1}_{X_u\leq(\log\varepsilon_K+N\log 2+N{\beta^2}/{2})/{\beta}}\big] \nonumber \\ & = \exp\bigg(\frac{\beta^2sN}{2}\bigg) \mathbb{E}\big[{\mathrm{e}}^{-\beta sX_u}\boldsymbol{1}_{X_u\leq(\log\varepsilon_K+N\log 2+N{\beta^2}/{2})/{\beta}}\big]. \end{align}
Note that
$X_u$
appearing in the expectation of (4.2) is distributed as a centered Gaussian variable with variance N, so the second expectation of (4.2) can be rewritten as
\begin{align} & \mathbb{E}\big[{\mathrm{e}}^{-\beta sX_u}\boldsymbol{1}_{X_u\leq(\log\varepsilon_K+N\log 2+N{\beta^2}/{2})/{\beta}}\big] \nonumber \\ & = \frac{1}{\sqrt{2\pi N}\,}\int_{-\infty}^{(\log\varepsilon_K+N\log 2+N{\beta^2}/{2})/{\beta}} {\mathrm{e}}^{-\beta sx}{\mathrm{e}}^{-{x^2}/{2N}}\,\mathrm{d}{x} \nonumber \\ & = {\mathrm{e}}^{{\beta^2s^2}N/{2}}\frac{1}{\sqrt{2\pi N}\,}\int_{-\infty}^{(\log\varepsilon_K+N\log 2+N{\beta^2}/{2})/{\beta}} {\mathrm{e}}^{-{(x+\beta s N)^2}/{2N}}\,\mathrm{d}{x} \nonumber \\ & = {\mathrm{e}}^{{\beta^2s^2}N/{2}}\frac{1}{\sqrt{2\pi N}\,}\int_{-(\log\varepsilon_K+N\log 2+N{\beta^2}/{2}+N\beta s)/{\beta\sqrt{N}}}^\infty {\mathrm{e}}^{-{y^2}/{2}}\,\mathrm{d}{y} \end{align}
\begin{align} = {\mathrm{e}}^{{\beta^2s^2}N/{2}}\frac{1}{\sqrt{2\pi N}\,}\int_{r(N)}^\infty{\mathrm{e}}^{-{y^2}/{2}}\,\mathrm{d}{y}, \end{align}
where (4.3) follows from the change of variables
$y=(x+\beta s)/\sqrt{N}$
and r(N) in (4.4) is defined as
$r(N)=-(\log \varepsilon_K + N\log 2 + N{\beta^2}/{2}+N \beta s)/{\beta\sqrt{N}}$
.
Recalling the definition of
$(\varepsilon_k)_{k=0}^K$
in the proof of Proposition 3.1, we have
\begin{align} \varepsilon_K = \frac{1}{2}\frac{1}{C^K}\prod_{n=0}^{K-1}\big((\eta_n)^{\gamma/2}-\eta_n\big)^{1/(10s)} \leq \prod_{n=0}^{K-1}\eta_n^{\gamma/(20s)} = \prod_{n=0}^{K-1}\eta_0^{\gamma^{n+1}/(20s)}, \end{align}
where the inequality follows from the fact that
$\gamma/2\geq 1$
,
$C\geq 1$
, and
$\eta_n\geq 0$
for all
$n\in[\![ 0,K-1]\!]$
. Recall that
$K=\lfloor 2\log_\gamma N \rfloor$
, so
\begin{equation*} \eqref{eq:eps bound} = \eta_0^{({\gamma}/{20s})(({\gamma^K-1})/({\gamma-1})} \leq \eta_0^{{\gamma N^2}/({20(\gamma-1)s})}. \end{equation*}
This and (4.5) imply that
\begin{align*} r(N) & \geq -\frac{1}{\beta\sqrt{N}\,}\bigg(\frac{\gamma}{20(\gamma-1)s}(\log\eta_0)N^2 + N\log 2 + N\frac{\beta^2}{2}+N \beta s\bigg) \\ & = c_1N^{3/2} - \bigg(\frac{\log 2}{\beta} + \frac{\beta}{2} + s\bigg) N^{1/2}, \end{align*}
and since
$\eta_0<1$
, we have
\begin{equation*} c_1 = -\frac{\gamma\log\!(\eta_0)}{20\beta(\gamma-1)s} >0. \end{equation*}
Therefore, there exists
$N_2=N_2(A,\beta,s)$
such that, for all
$N\geq N_2$
,
$r(N) \geq {c_1}N^{3/2}/{2}$
, and this implies that for
$N\geq N_2$
we have
\begin{equation} \int_{r(N)}^\infty{\mathrm{e}}^{-{y^2}/{2}}\,\mathrm{d}{y} \leq \int_{{c_1}N^{3/2}/{2}}^\infty{\mathrm{e}}^{-{x^2}/{2}}\,\mathrm{d}{x} \leq \frac{1}{(c_1/2)N^{3/2}}{\mathrm{e}}^{-{c_1^2N^3}/{8}}, \end{equation}
where the second inequality follows from a standard estimate for Gaussian tails (see, e.g., [Reference Durrett7, Theorem 1.2.6]). Combining (4.2), (4.4), and (4.6), we see that
\begin{multline*} \limsup_{N\rightarrow\infty}\mathbb{E}[(W_{\beta,N})^{-s}\boldsymbol{1}_{W_{\beta,N}}] \\ \leq \limsup_{N\rightarrow\infty}\bigg({\mathrm{e}}^{{\beta^2sN}/{2}}\cdot{\mathrm{e}}^{{\beta^2s^2}N/{2}}\frac{1}{\sqrt{2\pi N}\,} \cdot \frac{1}{(c_1/2) N^{3/2}}{\mathrm{e}}^{-{c_1^2N^3}/{8}}\bigg) = 0. \end{multline*}
From this and (4.1), we conclude that there exists
$N_0 = N_0(A,\beta,s)\geq \max\{N_1,N_2\}$
such that, for
$N\geq N_0$
,
$\mathbb{E}[(W_{\beta,N})^{-s}]\leq 2^s + C_1 + 1$
. By taking
$C = 2^s + C_1+1$
, the proof is completed.
Acknowledgements
I would like to thank Pascal Maillard for his guidance throughout the whole project and Michel Pain for his comments on the early draft of the paper. I also appreciate the constructive suggestions from an anonymous referee, which have substantially improved the quality of the paper.
Funding information
There are no funding bodies to thank relating to the creation of this article.
Competing interests
There were no competing interests to declare that arose during the preparation or publication process of this article.
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